Answer:
Explanation:
Given
mass of each car [tex]m=6.8\times 10^3 kg[/tex]
acceleration of train [tex]a=8\times 10^{-2} m/s^2[/tex]
Force required to pull this system
[tex]F=50\times ma[/tex]
For first car
[tex]F-T{1-2}=ma[/tex]
[tex]T_{1-2}=50ma-ma[/tex]
[tex]T_{1-2}=49 ma[/tex]
for second car
[tex]T_{1-2}-T_{2-3}=ma[/tex]
[tex]T_{2-3}=49 ma-ma[/tex]
[tex]T_{2-3}=48 ma[/tex]
this form a pattern of the form
[tex]T_{n-(n+1)}=(50-n)ma[/tex]
for (a)30 th and 31 st car tension
[tex]T_{30-31}=(50-30) ma[/tex]
[tex]T_{30-31}=20\times 6.8\times 10^{3}\times 8\times 10^{-2}[/tex]
[tex]T_{30-31}=10,880\ N[/tex]
For 49 th and 50 th car
[tex]T_{49-50}=(50-49) ma[/tex]
[tex]T_{49-50}=544\ N[/tex]
